Mathematics
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OpenStudy (adi3):
Will Medal
Please help!!
Use Quadratic formula to solve for x:
(x+2)(x-1) = 5
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OpenStudy (anonymous):
Start off by putting into \[ax^2+bx+c=0\] form
OpenStudy (adi3):
ok, wait let me solve it first
OpenStudy (anonymous):
Take your time
OpenStudy (adi3):
ok, x^2 +1x -7
OpenStudy (anonymous):
Good now we have the quadratic formula which is the following \[x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where a = 1, b= 1, and c = -7
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OpenStudy (adi3):
ok, wait again sorry
OpenStudy (adi3):
Answer is 2,-1
OpenStudy (anonymous):
Are you sure?
OpenStudy (adi3):
yes
OpenStudy (adi3):
no no i am wrong wait
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OpenStudy (adi3):
the answer is 2.15 and -3.15
OpenStudy (adi3):
i took the wrong equation
OpenStudy (anonymous):
\[x= \frac{ -1 \pm \sqrt{1^2-4(1)(-7)} }{ 2(1) } \implies x = \frac{ -1 \pm \sqrt{29} }{ 2 }\] correct?
OpenStudy (adi3):
yes
OpenStudy (anonymous):
That should \[x \approx -3.19,~~ \text{and}~~ x \approx 2.19\]
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OpenStudy (adi3):
can i ask you an another quetion pls
OpenStudy (anonymous):
Sure thing
OpenStudy (adi3):
Use Quadratic formula to solve for x:
\[x + \frac{ 1 }{ x+2 } = 4\]
OpenStudy (anonymous):
Same thing as earlier, try putting it into \[ax^2+bx+c=0\]
OpenStudy (adi3):
how will i do that
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OpenStudy (anonymous):
You can multiply through by (x+2)
OpenStudy (adi3):
??
OpenStudy (anonymous):
I mean you can find a common denominator which in this case is (x+2) if you like
OpenStudy (adi3):
can we cross multiply
OpenStudy (anonymous):
You can do as following \[\frac{ 1 }{ x+2 }=4-x \] now you may do as you like
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OpenStudy (anonymous):
\[\frac{ 1 }{ (x+2) } = (4-x)\]
OpenStudy (adi3):
ok
OpenStudy (adi3):
wait
OpenStudy (adi3):
is it = to 4x-x-5??
OpenStudy (anonymous):
Hmm, not quite, \[\frac{ 1 }{ (x+2) } = (4-x) \implies 1 = (4-x)(x+2)\]
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OpenStudy (adi3):
yeah then we multiply (4-x)(x+2) = 4x - x +6
OpenStudy (adi3):
an we bring it to the other side
OpenStudy (adi3):
-4x + x -6 + 1
OpenStudy (adi3):
= -4x + x -5 = 0
OpenStudy (adi3):
right??
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OpenStudy (anonymous):
Remember, we want our equation in \[ax^2+bx+c=0\] form, so what you wrote, does that make sense?
Think of what you did in the previous problem
OpenStudy (adi3):
x^2-4x-5
OpenStudy (adi3):
= 0
OpenStudy (anonymous):
\[(4-x)(x+2)=1 \implies 4x+8-x^2-2x =1 \implies 2x+8-x^2=1 \] so we can set this up as \[x^2-2x-7=0\]
OpenStudy (adi3):
ok. sorry i was offline, i had a lag
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OpenStudy (adi3):
answer is = 1.9, -0.9 right??
OpenStudy (adi3):
@iambatman
OpenStudy (anonymous):
Can you show how you got that?
OpenStudy (adi3):
ok |dw:1444558479751:dw|